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If p(x)=ax^4+bx^3+cx^2+dx+e has roots at x=1,2,3 and 4, and p(0)=48, what is p(5)
a)48
b)24
c)0
d)50
Read Solution (Total 1)
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- 48
if 1 is a root then x-1 is a factor of p(x)
similarly x-2 is a factor,x-3 ,x-4 are factors
but p(x) is 4th degree polynomial therefore it can be in the form
p(x) = k(x-1)(x-2)(x-3)(x-4) , where k is a constant
but given p(0) =48
therefore 24 k =48
k=2
p(x) =2(x-1)(x-2)(x-3)(x-4)
p(5)= 2*4*3*2*1 = 48 - 12 years agoHelpfull: Yes(43) No(3)
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